The Monty Hall problem is a counterintuitive probability puzzle named after the host of the American game show Let's Make a Deal. It poses a simple game: a contestant chooses one of three doors; behind one door is a valuable prize and behind the other two are lesser "goats" or booby prizes. The host, who knows what is behind every door, opens one of the remaining doors to reveal a goat, then offers the contestant the choice to stay with the original pick or switch to the other unopened door. Readers can find general background on probability at probability resources, the show at Let's Make a Deal, and the problem's namesake at Monty Hall.
Problem statement
In precise steps the game proceeds as follows: the host hides a single high-value prize behind one door and two low-value prizes behind the others; the contestant selects a door but does not open it; the host—aware of all door contents—opens one of the two remaining doors and always reveals a goat; finally, the contestant is invited to either keep the original choice or switch to the only other unopened door. For clarity, the toy model often describes the valuable item as a car and the decoys as goats; see a discussion of typical prize framing at prize examples and the decoys at goat examples. The central question is: does switching doors change the contestant's chance of winning the car?
Why switching helps
Many people's first intuition is that after one goat is revealed the two remaining doors each have an equal 50/50 chance. That impression ignores the host's knowledge and deliberate choice. A simple way to see the correct answer is to list the equally likely initial choices and track outcomes if the contestant always switches:
- Contestant initially picks the door with the car (probability 1/3). Host opens a goat door. If the contestant switches, they move to a goat and lose.
- Contestant initially picks goat A (probability 1/3). Host must open the other goat (if rules force revealing a goat), leaving the car behind the remaining closed door. Switching yields the car and a win.
- Contestant initially picks goat B (probability 1/3). By symmetry, switching again yields the car and a win.
Because two of the three initial possibilities lead to a car when switching, the switch strategy wins with probability 2/3, while staying wins with probability 1/3. This result hinges on the host always opening a door that he knows contains a goat and never opening the contestant's chosen door.
Another way to understand it is to note that the contestant's first choice has a 1/3 chance of being correct and a 2/3 chance of being wrong. When the host opens a goat door, he transfers no new probability to the chosen door; instead, the unopened, unchosen door inherits the 2/3 chance that the contestant's original pick was wrong. Computer simulations and repeated trials quickly confirm that switching wins roughly twice as often as staying.
History, controversy and variants
The puzzle gained widespread public attention in the early 1990s after being discussed in public columns and academic writing. The published solution provoked strong disagreement from many readers who found the 2/3 result surprising; subsequent explanations, visualizations and simulations helped settle the dispute for most observers. Variants of the problem explore what happens when the host's behavior is altered—for example, if the host sometimes opens the car door by mistake, if the host chooses a door at random when two goats are available, or if there are many more than three doors. Each change requires restating the host's rules explicitly: the 2/3 conclusion depends on the host deliberately avoiding the car and always offering the switch.
The Monty Hall problem is useful as a teaching example because it highlights conditional probability, the importance of information and how an agent's knowledge and actions affect outcomes. It also serves as a caution about trusting immediate intuition: carefully accounting for the process that generated the revealed information often changes the probabilities in non‑obvious ways.
